Determine the acceleration and normal force of system below. Assume no friction, the mass equal to 10.0 kg and the angle equal to 60.0 degrees. Assume gravity equals 9.8 m/s/s.
The picture looks like a right triangle with a box sliding down it.
I am a 9th grader in conceptual physics, so the calculus is already provided for us. So if you could solve this without calculus, that would be great!
The normal force of the box is mg*cosTheta.
The force of gravity down the plane is mg*SinTheta
so the net force down the plane is
forcegravity-friction
because the friction is zero, then you don't have to worry about normal force in the motion.
Netforce=mass*acceleration
or acceleration= mgSinTheta/m= g*sinTheta
To determine the acceleration and normal force of the system, let's break down the forces acting on the box.
First, consider the weight of the box. The weight can be calculated using the formula:
Weight = mass * gravity
Given that the mass of the box is 10.0 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the weight:
Weight = 10.0 kg * 9.8 m/s^2 = 98.0 N
Next, we need to resolve the weight force into components along the incline and perpendicular to it. Since the angle of the incline is given as 60 degrees, the weight force can be broken down as follows:
Weight (parallel to incline) = Weight * sin(angle)
Weight (perpendicular to incline) = Weight * cos(angle)
Plugging in the values, we have:
Weight (parallel to incline) = 98.0 N * sin(60 degrees) = 84.8 N
Weight (perpendicular to incline) = 98.0 N * cos(60 degrees) = 49.0 N
Now, let's consider the forces acting along the incline. The only force in this direction is the weight force component parallel to the incline, which acts in the downward direction. This force will cause the box to accelerate down the incline.
Using Newton's second law (F = ma) with the force parallel to the incline:
Sum of forces (parallel to incline) = Weight (parallel to incline) = ma
We can now solve for the acceleration:
a = Weight (parallel to incline) / mass = 84.8 N / 10.0 kg = 8.48 m/s^2
Therefore, the acceleration of the system is 8.48 m/s^2.
Finally, let's consider the normal force, which is the force perpendicular to the incline. In the absence of friction and any other vertical forces, the normal force will be equal in magnitude but opposite in direction to the perpendicular component of weight.
So, the normal force will be:
Normal force = Weight (perpendicular to incline) = 49.0 N
Therefore, the normal force acting on the box is 49.0 N.